Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) A voltaic cell is constructed with all reactants and products in their standard states. Will this condition hold as the cell operates? Explain. (b) Can the Nernst equation be used at temperatures other than room temperature? Explain. (c) What happens to the emf of a cell if the concentrations of the products are increased?

Short Answer

Expert verified
(a) No, the standard state condition will not hold as the cell operates. As reactants are consumed and products are generated, concentrations change, and the cell potential will decrease as the reaction progresses toward equilibrium. (b) Yes, the Nernst equation can be used at temperatures other than room temperature. Since the temperature appears explicitly in the equation, it can be used to calculate the cell potential at any given temperature by substituting the appropriate value for T. (c) As the product concentrations increase, the emf of a cell will decrease. This is because an increase in product concentrations leads to an increase in the reaction quotient Q, causing a decrease in cell potential (E) according to the Nernst equation.

Step by step solution

01

(a) Explanation of standard state condition in a voltaic cell

A standard state is defined as a situation where the concentration of each reactant and product is at 1 M, the pressure of gases is at 1 atm, and the temperature is 25°C (298 K). However, as the voltaic cell operates, the concentrations of reactants and products will change. Specifically, reactants will be consumed and products will be generated. Hence, the standard state condition will not hold as the cell operates. The cell potential will ultimately decrease as the reaction progresses toward equilibrium.
02

(b) Nernst equation and its applicability at different temperatures

The Nernst equation is a mathematical relationship that can be used to determine the cell potential at non-standard conditions. It is given by: \(E = E° - \frac{RT}{nF} \ln Q\) Where E is the cell potential, E° is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the redox reaction, F is the Faraday's constant, and Q is the reaction quotient. The Nernst equation can be used at temperatures other than room temperature (298 K). Since the temperature appears explicitly in the equation, it can be used to calculate the cell potential at any given temperature by substituting the appropriate value for T.
03

(c) Effect of product concentration on the emf of a cell

As the product concentrations increase, the reaction quotient (Q) in the Nernst equation will also increase. According to the Nernst equation: \(E = E° - \frac{RT}{nF} \ln Q\) An increase in Q will lead to an increase in the natural logarithm term (\(\ln Q\)), which in turn will increase the entire term\(\frac{RT}{nF} \ln Q\). Since this term is subtracted from the standard cell potential (E°), an increase in the product concentrations will result in a decrease in the cell potential (E). Therefore, the emf of a cell will decrease when the concentrations of the products are increased.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Write the anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II). (b) Write the balanced half-reactions involved in the air oxidation of \(\mathrm{Fe}^{2+}(a q)\) to \(\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}\)

A voltaic cell utilizes the following reaction and operates at \(298 \mathrm{~K}\) : $$ 3 \mathrm{Ce}^{4+}(a q)+\mathrm{Cr}(s)-\rightarrow 3 \mathrm{Ce}^{3+}(a q)+\mathrm{Cr}^{3+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ce}^{4+}\right]=3.0 M\), \(\left[\mathrm{Ce}^{3+}\right]=0.10 \mathrm{M}\), and \(\left[\mathrm{Cr}^{3+}\right]=0.010 \mathrm{M} ?\) (c) What is the emf of the cell when \(\left[\mathrm{Ce}^{4+}\right]=0.10 \mathrm{M},\left[\mathrm{Ce}^{3+}\right]=1.75 \mathrm{M}\), and \(\left[\mathrm{Cr}^{3+}\right]=2.5 \mathrm{M} ?\)

In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case. (a) \(\mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g)-\mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow\) \(2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)\) (c) \(3 \mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)-\cdots\) \(3 \mathrm{~S}(s)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q)-\mathrm{-} \rightarrow\) \(\mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

(a) What is meant by the term oxidation? (b) On which side of an oxidation half-reaction do the electrons appear? (c) What is meant by the term oxidant? (d) What is meant by the term oxidizing agent?

Is each of the following substances likely to serve as an oxidant or a reductant: (a) \(\mathrm{Ce}^{3+}(a q)\), (b) \(\mathrm{Ca}(\mathrm{s})\), (c) \(\mathrm{ClO}_{3}^{-}(a q)\), (d) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) ?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free